# Tag Info

2

The standard construction of the real numbers via Cauchy sequences can be understood algebraically as follows: Let $R$ be the ring of Cauchy sequences of rational numbers under the operations of component-wise addition and multiplication. Then $\mathfrak m = \{(x_n)_{n \in \mathbb N} \in R \mid x_n \to 0\}$ is a maximal ideal of $R$, and we define $\mathbb R ... 1 You have:$\sqrt 2 := \dfrac{2}{\frac{2}{\frac{2}{\frac{2}{\vdots\over\sqrt 2}}}} = \dfrac{2}{2}\cdot\frac{2}{2}\cdots\sqrt 2$Which is not a very helpful definition. 5 The problem is that if you take$a_0$non-zero and you define$a_n = 2/a_{n-1}$, you have the relation$a_n = a_{n-2}$(check that) so that the sequence converges if and only if$a_n = a_{n-1}$, which implies$a_0 = \pm \sqrt{2}$. So we cannot "deduce by analogy" because there is no analogy to make. The difference with the continued fractions is that in this ... 2 It is quite "natural" that early convergences are "problematic". In fact,$S_{-1}$is always$\frac{1}{0}$. (Sometimes people define$p_{-2}=0$and$q_{-2}=1$, so$S_{-2} \equiv 0$)$q_1=0$is not a big problem, since you notice that$q_2 = a_2q_1+q_0 = 1 \cdot 0 + 1 = 1$and all following$q_k$are good. Fact:$S_k$is a better approximation than$S_n ...

2

Proposition 19 says we need to check a few things: $141 > (\pm 4)^2 + \frac{1}{2}(|\pm 4| + 1)$ Then $x^2 - 141y^2 = \pm 4$ appears as solution to continued fraction expansion to $\sqrt{141}$. Since $\pm 4$ does not appear in your remainders, $x^2 - 141 y^2 = \pm 4$ does not have a solution. Therefore you just solve Pell equation with $r = 1$, which ...

2

(Not an answer but too long for a comment.) Here is another "bizarre" one involving $e$ and $\pi$, $$\sqrt{\frac{\pi\,e^x}{2x}}=1+\frac{x}{1\cdot3}+\frac{x^2}{1\cdot3\cdot5}+\frac{x^3}{1\cdot3\cdot5\cdot7}+\dots+\cfrac1{1+\cfrac{1}{x+\cfrac{2}{1+\cfrac{3}{x+\ddots}}}}$$ No need to mention who found this. As Kevin Brown of Mathpages commented in this old ...

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$\sqrt{141}=[11,\overline{1,6,1,22}]$. Yes $95+8\sqrt{141}$ is a fundamental unit.

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For any $D>0$ there are only finitely many reduced elements $\beta$ whose minimal polynomial has discriminant $D$ The proof relies on the fact that there are only finitely many possible values for the coefficients $A,B$ in the minimal polynomial $A x^2+B x+C$ of $\beta$ Apply this result to $\mathbb{Q}\left(\sqrt{30}\right)$ Now $D=30m^2$ for some ...

3

If I understand the definition correctly, all you need to do is solve the system of inequalities $$a+b\sqrt{30}>1\text{ and } 0>a-b\sqrt{30}>-1$$ for rational numbers. You can't of course expect to list the solutions -there are infinitely many. It is not difficult to describe the set though: The first inequality is a half plane; to see that ...

1

We have that $\frac{\sqrt{7}}{3}$ is positive but less than one, hence its continued fraction starts with a $\color{red}{0}$; $\frac{3}{\sqrt{7}}=\frac{3\sqrt{7}}{7}=1+\frac{3\sqrt{7}-7}{7}$, hence the next element of the continued fraction is $\color{red}{1}$; $\frac{7}{3\sqrt{7}-7}=\frac{3\sqrt{7}+7}{2}=7+\frac{3\sqrt{7}-7}{2}$, hence the next element of ...

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